A brief solution to the written exam in STK9900/STK4900 Statistical ...

A brief solution to the written exam in STK9900/STK4900
Statistical methods and applications.
Thursday 6. June 2013.
The exam in STK4900 and STK9900 have substantial overlap, but are not the same. This
is a short solution to the STK9900 questions, which includes the questions for STK4900.
Exercise 1
a) To test if the expression of gene 1 is a significant explanatory variable for bone density,
we use the t-test statistic
t = ˆβ 1
se( ˆβ 1 )
where ˆβ 1 is the estimated effect of gene 1 and se( ˆβ 1 ) is the corresponding standard error.
Using the output in (I), the test statistic takes the value
t = −1.1297
0.3609 = −3.13.
The value -3.13 is to be compared with the t-distribution with 82 degrees of freedom.
Using the table for t-distributions, we find that the P-value is smaller than 2 · 0.005 = 0.01
(two-sided, using 60 degrees of freedom from the table). Hence β 1 is significantly different
from 0 and gene 1 is a significant explanatory variable for bone density.
b) In the results (II) we see that gene 1 is no longer significant when gene 2 is included.
Gene 2 i highly significant. Hence if we have information on gene 2, gene 1 does not contain
enough additional information on bone density to be significant. This can happen when
the covariates are (highly) correlated. We see that the correlation between gene 1 and gene
2 is 0.73.
c) The second model has only significant covariates and a slightly higher Adjusted R 2 ,
hence should be considered better that the first (the first model has a higher R 2 , but has
also one more covariate, so we use Adjusted R 2 for comparison here). For the second model,
we interpret the estimated effects as: A high gene expression for gene 2 tends to reduce
expected bone density, when the other covariates (genes) are kept unchanged. A high gene
expression for gene 3 tends to increase expected bone density, when the other genes are
unchanged. The same for gene 4. We also see that gene 3 has the largest estimated effect
on bone density.
d) [9900] Short answer: Interaction between covariates is when the effect of one covariate
on the response depends on the level of another covariate. Some more discussion appreciated,
explaining how this can be modelled etc., see lecture notes Lecture 4. In the setting
here, interaction between genes would mean f.ex. that the effect of a high expression of
gene r could depend on the expression level of another gene s.
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Exercise 2
a) We have data (x 1 , x 2 , y) for n = 35 subjects (surgeries), where x 1 and x 2 are explanatory
vaiables and the outcome y is binary (0 and 1). We are interested in modelling the
probability of getting a sore throat as a function of the explanatory variables, hence we
model p(x 1 , x 2 ) = P (y = 1|x 1 , x 2 ) with a logistic regression model
p(x 1 , x 2 ) = exp(β 0 + β 1 x 1 + β 2 x 2 )
1 + exp(β 0 + β 1 x 1 + β 2 x 2 ) .
This ensures that p(x 1 + x 2 ) is a proper probability, and the log odds will be linear in the
covariates.
b) With only one explanatory variable x 1 , we find
estimated as
P (y = 1|x 1 = 30) = exp(β 0 + β 1 · 30)
1 + exp(β 0 + β 1 · 30)
exp(−2.21358 + 0.07038 · 30)
1 + exp(−2.21358 + 0.07038 · 30) = 0.4745.
c) The odds ratio for 40 minutes wrt. 30 minutes is exp(10 · ˆβ 1 ) = exp(0.7038) = 2.02 .
A confidence interval for 10 · β 1 is found through
10 · ˆβ 1 ± 1.96se(10 · ˆβ 1 ),
where se(10· ˆβ 1 ) = 10·se( ˆβ 1 ). With the estimates from R we get 0.7038±1.96·0.2667, giving
the 95% confidence interval (0.181068, 1.226532). This gives a 95% confidence interval for
the odds ratio as
(exp(0.181068), exp(1.226532))
that is, (1.199, 3.409). The OR tells us that the odds for a sore throat are doubled if the
surgery lasts 10 minutes more. From the confidence interval we can conclude that there is
some uncertainty about this doubling, but at least we can say that the odds are increased
(starting above 1).
d) Wald test:
H 0 : β 2 = 0 versus H a : β 2 ≠ 0
From R we find the z-value -1.798, and a P-value= 2P (Z > 1.798) = 2 · 0.036 = 0.072
where Z is standard normal. This P-value is not small enough to reject H 0 at level 0.05,
so the type (x 2 ) is not significantly improving the model.
Deviance test:
G = D 0 − D = 33.651 − 30.138 = 3.513
G is χ 2 with 1 df under H 0 . From the χ 2 table we find that the P-value must be larger
than 0.05, and conclude again that type (x 2 ) is not significantly improving the model.
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Exercise 3
a) The estimated rate ratio is exp(1.2016) = 3.325, hence the rate of wave caused damages
is more than three times larger for the ships built in 70-74 compared to 60-64.
b) [9900] For an explanation of over dispersion, see lecture notes for Lecture 8. Should
write down the overdispersed Poisson regression model, introducing V ar(Y i ) = φµ i , and
explain that using the quasi family in the glm allows to estimate φ and corrects the standard
errors to
√
se ∗ = se ˆφ.
For the ship damage data, ˆφ = 3.19643 and the standard errors of the estimated coefficients
are scaled up with a factor
√
ˆφ = 1.788,
and the test statistics and p-values corrected correspondingly. The p-values become bigger,
but the effects are still significant.
c) This is a Poisson model with observations that are aggregated counts. If Y i is the
number of counts in group i, the model can be written as
E(Y i ) = exp(log(months i ) + β 0 + β 1 x Bi + β 2 x Ci + β 3 x Di + β 4 x Ei
+β 5 x 65−69,i + β 6 x 70−74,i + β 7 x 75−79,i + β 8 x op75−79,i )
where x Bi = 1 if ship type in group i is B and 0 otherwise, etc. We test
using a deviance test:
H 0 : β 1 = β 2 = β 3 = β 4 = β 8 = 0
G = D 0 − D = 73.225 − 38.695 = 34.53.
G is χ 2 with 5 df under H 0 . P (χ 2 5 > 34.53) < 0.005 and we reject H 0 at any reasonable
level of significance. Hence we can conclude that the new variables significantly improve
the model.
d) Using estimated rate ratios, we find that ship type B, C and D all are safer than ship
type A, when all other covariates are unchanged. Ship type E is more dangerous.
Specifically we find RR=exp(-0.6874)=0.5029 for ship type C compared to ship type A,
when all other covariates are unchanged. Similarly we find RR=exp(-0.5433)=0.5808 for
ship type B compared to A. Ship type C has therefore the lowest rate of wave caused
damage, but the standard error of ˆβ 2 is much larger than that of ˆβ 1 , so it is also reasonable
to argue that ship type B should be considered the safest, given that the difference in RR
is very small.
The RR for operation in the last period wrt the first is RR=exp(0.38447)=1.4688. This
means that there was an increased risk of damage for operations in the period 1975-79
compared to the previous period 1960-74, when all other covariates as ship type and year
of construction are the same.
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